Integrand size = 22, antiderivative size = 110 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=\frac {x}{c^3}+\frac {1}{12 a c^3 (1-a x)^3}-\frac {5}{8 a c^3 (1-a x)^2}+\frac {39}{16 a c^3 (1-a x)}-\frac {1}{16 a c^3 (1+a x)}+\frac {9 \log (1-a x)}{4 a c^3}-\frac {\log (1+a x)}{4 a c^3} \] Output:
x/c^3+1/12/a/c^3/(-a*x+1)^3-5/8/a/c^3/(-a*x+1)^2+39/16/a/c^3/(-a*x+1)-1/16 /a/c^3/(a*x+1)+9/4*ln(-a*x+1)/a/c^3-1/4*ln(a*x+1)/a/c^3
Time = 0.09 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.75 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=\frac {\frac {2 \left (-11+7 a x+24 a^2 x^2-15 a^3 x^3-12 a^4 x^4+6 a^5 x^5\right )}{(-1+a x)^3 (1+a x)}+27 \log (1-a x)-3 \log (1+a x)}{12 a c^3} \] Input:
Integrate[E^(2*ArcCoth[a*x])/(c - c/(a^2*x^2))^3,x]
Output:
((2*(-11 + 7*a*x + 24*a^2*x^2 - 15*a^3*x^3 - 12*a^4*x^4 + 6*a^5*x^5))/((-1 + a*x)^3*(1 + a*x)) + 27*Log[1 - a*x] - 3*Log[1 + a*x])/(12*a*c^3)
Time = 0.85 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6717, 27, 6707, 6700, 99, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx\) |
\(\Big \downarrow \) 6717 |
\(\displaystyle -\int \frac {a^6 e^{2 \text {arctanh}(a x)}}{c^3 \left (a^2-\frac {1}{x^2}\right )^3}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a^6 \int \frac {e^{2 \text {arctanh}(a x)}}{\left (a^2-\frac {1}{x^2}\right )^3}dx}{c^3}\) |
\(\Big \downarrow \) 6707 |
\(\displaystyle \frac {a^6 \int \frac {e^{2 \text {arctanh}(a x)} x^6}{\left (1-a^2 x^2\right )^3}dx}{c^3}\) |
\(\Big \downarrow \) 6700 |
\(\displaystyle \frac {a^6 \int \frac {x^6}{(1-a x)^4 (a x+1)^2}dx}{c^3}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {a^6 \int \left (-\frac {1}{4 a^6 (a x+1)}+\frac {1}{16 a^6 (a x+1)^2}+\frac {1}{a^6}+\frac {9}{4 a^6 (a x-1)}+\frac {39}{16 a^6 (a x-1)^2}+\frac {5}{4 a^6 (a x-1)^3}+\frac {1}{4 a^6 (a x-1)^4}\right )dx}{c^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^6 \left (\frac {39}{16 a^7 (1-a x)}-\frac {1}{16 a^7 (a x+1)}-\frac {5}{8 a^7 (1-a x)^2}+\frac {1}{12 a^7 (1-a x)^3}+\frac {9 \log (1-a x)}{4 a^7}-\frac {\log (a x+1)}{4 a^7}+\frac {x}{a^6}\right )}{c^3}\) |
Input:
Int[E^(2*ArcCoth[a*x])/(c - c/(a^2*x^2))^3,x]
Output:
(a^6*(x/a^6 + 1/(12*a^7*(1 - a*x)^3) - 5/(8*a^7*(1 - a*x)^2) + 39/(16*a^7* (1 - a*x)) - 1/(16*a^7*(1 + a*x)) + (9*Log[1 - a*x])/(4*a^7) - Log[1 + a*x ]/(4*a^7)))/c^3
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x _Symbol] :> Simp[c^p Int[x^m*(1 - a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p] || GtQ[c, 0])
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symb ol] :> Simp[d^p Int[(u/x^(2*p))*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x ] /; FreeQ[{a, c, d, n}, x] && EqQ[c + a^2*d, 0] && IntegerQ[p]
Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Simp[(-1)^(n/2) Int[ u*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[a, x] && IntegerQ[n/2]
Time = 0.17 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.76
method | result | size |
default | \(\frac {a^{6} \left (\frac {x}{a^{6}}-\frac {\ln \left (a x +1\right )}{4 a^{7}}-\frac {1}{16 a^{7} \left (a x +1\right )}-\frac {1}{12 a^{7} \left (a x -1\right )^{3}}-\frac {5}{8 a^{7} \left (a x -1\right )^{2}}-\frac {39}{16 a^{7} \left (a x -1\right )}+\frac {9 \ln \left (a x -1\right )}{4 a^{7}}\right )}{c^{3}}\) | \(84\) |
risch | \(\frac {x}{c^{3}}+\frac {-\frac {5 a^{2} c^{3} x^{3}}{2}+2 a \,c^{3} x^{2}+\frac {13 c^{3} x}{6}-\frac {11 c^{3}}{6 a}}{c^{6} \left (a x -1\right )^{2} \left (a^{2} x^{2}-1\right )}+\frac {9 \ln \left (-a x +1\right )}{4 a \,c^{3}}-\frac {\ln \left (a x +1\right )}{4 a \,c^{3}}\) | \(93\) |
norman | \(\frac {\frac {a^{5} x^{6}}{c}-\frac {5 x}{2 c}+\frac {3 a \,x^{2}}{2 c}+\frac {31 a^{2} x^{3}}{6 c}-\frac {8 a^{3} x^{4}}{3 c}-\frac {17 a^{4} x^{5}}{6 c}}{\left (a x -1\right )^{3} c^{2} \left (a x +1\right )^{2}}+\frac {9 \ln \left (a x -1\right )}{4 a \,c^{3}}-\frac {\ln \left (a x +1\right )}{4 a \,c^{3}}\) | \(107\) |
parallelrisch | \(\frac {12 a^{5} x^{5}+27 \ln \left (a x -1\right ) x^{4} a^{4}-3 \ln \left (a x +1\right ) x^{4} a^{4}-46 a^{4} x^{4}-54 a^{3} \ln \left (a x -1\right ) x^{3}+6 \ln \left (a x +1\right ) x^{3} a^{3}+14 a^{3} x^{3}+48 a^{2} x^{2}+54 a \ln \left (a x -1\right ) x -6 \ln \left (a x +1\right ) x a -30 a x -27 \ln \left (a x -1\right )+3 \ln \left (a x +1\right )}{12 c^{3} \left (a x -1\right )^{2} \left (a^{2} x^{2}-1\right ) a}\) | \(156\) |
Input:
int(1/(a*x-1)*(a*x+1)/(c-c/a^2/x^2)^3,x,method=_RETURNVERBOSE)
Output:
a^6/c^3*(x/a^6-1/4*ln(a*x+1)/a^7-1/16/a^7/(a*x+1)-1/12/a^7/(a*x-1)^3-5/8/a ^7/(a*x-1)^2-39/16/a^7/(a*x-1)+9/4/a^7*ln(a*x-1))
Time = 0.10 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.25 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=\frac {12 \, a^{5} x^{5} - 24 \, a^{4} x^{4} - 30 \, a^{3} x^{3} + 48 \, a^{2} x^{2} + 14 \, a x - 3 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (a x + 1\right ) + 27 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (a x - 1\right ) - 22}{12 \, {\left (a^{5} c^{3} x^{4} - 2 \, a^{4} c^{3} x^{3} + 2 \, a^{2} c^{3} x - a c^{3}\right )}} \] Input:
integrate(1/(a*x-1)*(a*x+1)/(c-c/a^2/x^2)^3,x, algorithm="fricas")
Output:
1/12*(12*a^5*x^5 - 24*a^4*x^4 - 30*a^3*x^3 + 48*a^2*x^2 + 14*a*x - 3*(a^4* x^4 - 2*a^3*x^3 + 2*a*x - 1)*log(a*x + 1) + 27*(a^4*x^4 - 2*a^3*x^3 + 2*a* x - 1)*log(a*x - 1) - 22)/(a^5*c^3*x^4 - 2*a^4*c^3*x^3 + 2*a^2*c^3*x - a*c ^3)
Time = 0.34 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.93 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=a^{6} \left (\frac {- 15 a^{3} x^{3} + 12 a^{2} x^{2} + 13 a x - 11}{6 a^{11} c^{3} x^{4} - 12 a^{10} c^{3} x^{3} + 12 a^{8} c^{3} x - 6 a^{7} c^{3}} + \frac {x}{a^{6} c^{3}} + \frac {\frac {9 \log {\left (x - \frac {1}{a} \right )}}{4} - \frac {\log {\left (x + \frac {1}{a} \right )}}{4}}{a^{7} c^{3}}\right ) \] Input:
integrate(1/(a*x-1)*(a*x+1)/(c-c/a**2/x**2)**3,x)
Output:
a**6*((-15*a**3*x**3 + 12*a**2*x**2 + 13*a*x - 11)/(6*a**11*c**3*x**4 - 12 *a**10*c**3*x**3 + 12*a**8*c**3*x - 6*a**7*c**3) + x/(a**6*c**3) + (9*log( x - 1/a)/4 - log(x + 1/a)/4)/(a**7*c**3))
Time = 0.03 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.88 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=-\frac {15 \, a^{3} x^{3} - 12 \, a^{2} x^{2} - 13 \, a x + 11}{6 \, {\left (a^{5} c^{3} x^{4} - 2 \, a^{4} c^{3} x^{3} + 2 \, a^{2} c^{3} x - a c^{3}\right )}} + \frac {x}{c^{3}} - \frac {\log \left (a x + 1\right )}{4 \, a c^{3}} + \frac {9 \, \log \left (a x - 1\right )}{4 \, a c^{3}} \] Input:
integrate(1/(a*x-1)*(a*x+1)/(c-c/a^2/x^2)^3,x, algorithm="maxima")
Output:
-1/6*(15*a^3*x^3 - 12*a^2*x^2 - 13*a*x + 11)/(a^5*c^3*x^4 - 2*a^4*c^3*x^3 + 2*a^2*c^3*x - a*c^3) + x/c^3 - 1/4*log(a*x + 1)/(a*c^3) + 9/4*log(a*x - 1)/(a*c^3)
Time = 0.12 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.73 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=\frac {x}{c^{3}} - \frac {\log \left ({\left | a x + 1 \right |}\right )}{4 \, a c^{3}} + \frac {9 \, \log \left ({\left | a x - 1 \right |}\right )}{4 \, a c^{3}} - \frac {15 \, a^{3} x^{3} - 12 \, a^{2} x^{2} - 13 \, a x + 11}{6 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{3} a c^{3}} \] Input:
integrate(1/(a*x-1)*(a*x+1)/(c-c/a^2/x^2)^3,x, algorithm="giac")
Output:
x/c^3 - 1/4*log(abs(a*x + 1))/(a*c^3) + 9/4*log(abs(a*x - 1))/(a*c^3) - 1/ 6*(15*a^3*x^3 - 12*a^2*x^2 - 13*a*x + 11)/((a*x + 1)*(a*x - 1)^3*a*c^3)
Time = 0.06 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.85 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=\frac {x}{c^3}-\frac {\frac {13\,x}{6}+2\,a\,x^2-\frac {11}{6\,a}-\frac {5\,a^2\,x^3}{2}}{-a^4\,c^3\,x^4+2\,a^3\,c^3\,x^3-2\,a\,c^3\,x+c^3}+\frac {9\,\ln \left (a\,x-1\right )}{4\,a\,c^3}-\frac {\ln \left (a\,x+1\right )}{4\,a\,c^3} \] Input:
int((a*x + 1)/((c - c/(a^2*x^2))^3*(a*x - 1)),x)
Output:
x/c^3 - ((13*x)/6 + 2*a*x^2 - 11/(6*a) - (5*a^2*x^3)/2)/(c^3 + 2*a^3*c^3*x ^3 - a^4*c^3*x^4 - 2*a*c^3*x) + (9*log(a*x - 1))/(4*a*c^3) - log(a*x + 1)/ (4*a*c^3)
Time = 0.17 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.39 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx=\frac {27 \,\mathrm {log}\left (a x -1\right ) a^{4} x^{4}-54 \,\mathrm {log}\left (a x -1\right ) a^{3} x^{3}+54 \,\mathrm {log}\left (a x -1\right ) a x -27 \,\mathrm {log}\left (a x -1\right )-3 \,\mathrm {log}\left (a x +1\right ) a^{4} x^{4}+6 \,\mathrm {log}\left (a x +1\right ) a^{3} x^{3}-6 \,\mathrm {log}\left (a x +1\right ) a x +3 \,\mathrm {log}\left (a x +1\right )+12 a^{5} x^{5}-39 a^{4} x^{4}+48 a^{2} x^{2}-16 a x -7}{12 a \,c^{3} \left (a^{4} x^{4}-2 a^{3} x^{3}+2 a x -1\right )} \] Input:
int(1/(a*x-1)*(a*x+1)/(c-c/a^2/x^2)^3,x)
Output:
(27*log(a*x - 1)*a**4*x**4 - 54*log(a*x - 1)*a**3*x**3 + 54*log(a*x - 1)*a *x - 27*log(a*x - 1) - 3*log(a*x + 1)*a**4*x**4 + 6*log(a*x + 1)*a**3*x**3 - 6*log(a*x + 1)*a*x + 3*log(a*x + 1) + 12*a**5*x**5 - 39*a**4*x**4 + 48* a**2*x**2 - 16*a*x - 7)/(12*a*c**3*(a**4*x**4 - 2*a**3*x**3 + 2*a*x - 1))